Three particle of mas m each are placed at the three corners of an equlateral triangle of side a. Find the work which should be done on this system to increase the sides of the triangle to 2a.

Explanation

To solve the problem of finding the work done on a system of three particles of mass \( m \) each, placed at the corners of an equilateral triangle of side \( a \), when the side is increased to \( 2a \), we can follow these steps:

### Step 1: Calculate the Initial Potential Energy
The gravitational potential energy \( U \) between two masses \( m \) and \( m \) separated by a distance \( r \) is given by:

\[ U = -\frac{G m^2}{r} \]

For three particles at the corners of an equilateral triangle with side length \( a \), the potential energy between each pair of particles is:

1. Between particle 1 and particle 2: \( U_{12} = -\frac{G m^2}{a} \)
2. Between particle 1 and particle 3: \( U_{13} = -\frac{G m^2}{a} \)
3. Between particle 2 and particle 3: \( U_{23} = -\frac{G m^2}{a} \)

The total initial potential energy \( U_{\text{initial}} \) is:

\[ U_{\text{initial}} = U_{12} + U_{13} + U_{23} = 3 \left(-\frac{G m^2}{a}\right) = -\frac{3G m^2}{a} \]

### Step 2: Calculate the Final Potential Energy
When the side length of the triangle is increased to \( 2a \), the potential energy between each pair of particles becomes:

1. Between particle 1 and particle 2: \( U_{12} = -\frac{G m^2}{2a} \)
2. Between particle 1 and particle 3: \( U_{13} = -\frac{G m^2}{2a} \)
3. Between particle 2 and particle 3: \( U_{23} = -\frac{G m^2}{2a} \)

The total final potential energy \( U_{\text{final}} \) is:

\[ U_{\text{final}} = U_{12} + U_{13} + U_{23} = 3 \left(-\frac{G m^2}{2a}\right) = -\frac{3G m^2}{2a} \]

### Step 3: Calculate the Work Done
The work done \( W \) by the external agent to increase the distance between the particles is equal to the change in potential energy:

\[ W = U_{\text{final}} - U_{\text{initial}} \]

Substituting the values we found:

\[ W = \left(-\frac{3G m^2}{2a}\right) - \left(-\frac{3G m^2}{a}\right) \]

This simplifies to:

\[ W = -\frac{3G m^2}{2a} + \frac{3G m^2}{a} = -\frac{3G m^2}{2a} + \frac{6G m^2}{2a} = \frac{3G m^2}{2a} \]

### Final Answer
The work done on the system to increase the sides of the triangle from \( a \) to \( 2a \) is:

\[ W = \frac{3G m^2}{2a} \]